What is the sum from n=0 to infinity of 1/(1-6n) ?

The sum from n=0 to infinity of 1/(1-6n) is diverges

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sum from n=0 to infinity of 1/(1-6n)

n = 0 \sum \infty \frac{1}{1 - 6 n}

diverges

Solution steps

n = 0 \sum \infty \frac{1}{1 - 6 n}

Factor

1 - 6 n : - (- 1 + 6 n)

= n = 0 \sum \infty \frac{1}{- ( - 1 + 6 n )}

Apply the constant multiplication rule:

\sum c \cdot a_{n} = c \cdot \sum a_{n}

= - n = 0 \sum \infty \frac{1}{- 1 + 6 n}

Apply Series Integral Test:

diverges

= - diverges

= diverges

n = 0 \sum \infty (0.2)^{n}

n = 0 \sum \infty \frac{n ^{2}}{9 ^{n}}

n = 0 \sum \infty n \cdot e^{- n}

n = 0 \sum \infty \frac{( - 4 ) ^{n}}{n}

n = 0 \sum \infty \frac{1}{5 ^{n} - 4 ^{n}}

What is the sum from n=0 to infinity of 1/(1-6n) ?
The sum from n=0 to infinity of 1/(1-6n) is diverges